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which, after interchanging rows and columns and then multiplying by 1=D (which in this case is 1=1 1), we have ! 5 7 A 1 ; answer: 3 4 233. First,  2 4   D 6 0 2   0 5  2     2  1  6 2   5  3
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 1  132 3
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Next, in the manner of eq. (483), the value of A0 is
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Note that, in this particular example, the values of the determinants are all easy to nd because each can be expanded in terms of a row or column in which all the elements except one are zero. (Note that three of the determinants have the value zero by inspection, since, if the elements of any row or column are all equal to zero, the value of the determinant is zero.) You can now verify that the above matrix reduces to the form
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thus 2 6 0 6 A0 6 4 12 48 66 0 33 0 22 36 6 12 24 07 7 7 60 5 24 0 3
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Now, in the last expression, interchange the rows and columns. Doing this, and remembering to multiply the result by 1=D, we have that 3 2 66 0 12 48 0 36 12 7 1 6 0 7 6 A 1 7; answer: 6 132 4 33 22 6 24 5 0 0 60 24
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234. First (we ll expand in terms of the elements of the second row) we have        8 1   3 8 0  60 D 2  6 4    6 3 Then
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therefore, 2 A 1 6 4 2 3 2 1=10 1=15 3=10 8=15 1=30 3
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1 6 18 4 18 60 12 32
7 6 24 5 4 3=10 1=5 16
7 4=10 5 4=15
either way is a correct answer. 235. First, in the manner of eq. (476), write the rst set of equations in the matrix form 2 3 4 1 6 32 3 2 3 r x 76 7 6 7 5 54 y 5 4 s 5 t z 2 1
6 4 2 4
In the above, now verify that D 164. Then, in the manner of eq. (480), eq. (A) can be written as follows, where the superscript 1 indicates that the inverse of the
matrix is to be taken 2 3 2 x 3 4 6 7 6 1 4 y 5 4 2 z 4 6
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3 1 2 3 r 1 7 6 7 5 5 4 s 5 2 t
To nd the indicated inverse of the matrix, let us next nd the value of A0 which is, in this case, using eq. (483), 2 3 28 24 16 7 6 A0 4 2 10 34 5 19 13 5 and thus 2 3 4 1 3 1 2 28 2 10 34 19 3
6 4 2 4
1 6 7 1 5 5 4 24 164 6 2 16
7 13 5 5
therefore, eq. (B) becomes 2 2 3 x 28 1 6 6 7 4 24 4y5 164 z 16 32 3 r 76 7 10 13 54 s 5 t 34 5 2 19
Now take the product of the two matrices on the right-hand side, as indicated; the right-hand side then becomes a 3 1 matrix; thus 2 3 2 3 x 28r 2s 19t 1 6 6 7 7 4y5 4 24r 10s 13t 5 164 z 16r 34s 5t Now carry out the indicated multiplication of the right-hand side by (1/164) (section 11.1). We then have the equality of two 3 1 matrices, and therefore, by the de nition of equal matrices from section 11.1, the last matrix equation is the equivalent of the following three simultaneous equations 28=164 r 2=164 s 19=164 t x 24=164 r 10=164 s 13=164 t y 16=164 r 34=164 s 5=164 t z Now compare the coe cients in the last three equations with the corresponding ones in set 2; doing this, you ll nd that a 28=164 , b 2=164 , c 19=164 , and so on, to the nal value, i 5=164 , answers.
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